Question

Calculate %m/v composition of 0.022 Kg ammonium nitrate in 587g solution (d=1.07 g/mL)

Answer 1

`V_(tot) = (587 g)/(1,07 g/mL) = 549 mL`

0,022 kg = 22 g

`(m)/(V) = (22 g × 100)/(549 mL) = 4,0 %`